Question

PLEASE HELP: Two Masses, a Pulley, and an Inclined Plane

Block 1, of mass m1 = 0.700kg , is connected over an ideal (massless and frictionless) pulley to block 2, of mass m2, as shown. For an angle of θ = 30.0∘ and a coefficient of kinetic friction between block 2 and the plane of μ = 0.200, an acceleration of magnitude a = 0.200m/s2 is observed for block 2.

find the mass of block 2, m2:

Answer 1

I thought you would use this equation:
\[\frac{ m _{1} }{ m _{2} }= \frac{ a+g \mu \cos (\theta) +g \sin (\theta) }{ g-a}\]

Answer 2

or....

Answer 3

i would prefer to use the above equation

Answer 4

I tired it, it is not the right answer

Answer 5

i would prefer writing eqns of motion for both blocks. For that ,i need to know the direction of acceleration of m2 whose magnitude is 0.2m/s2 (given). It has to be upwards because friction is acting downwards in the figure and friction and resultant acc always point in opposite direction.

Answer 6

okay, that makes sense

Answer 7

Now ,we can write eqns of motion as follows. Noting that m2 is accelerated upwards the plane, therefore m1 will be accelerated down with same acceleration ,0.2m/s2.

Forces on m1 are :

T -upwards
m1g- downwards
Since acceleration, 0.2 of m1 is downwards ,so
m1a = m1g-T where a =0.2m/s/s ................................(1)

Forces on m2 are:

T -upwards the plane
m2*g*sin30 -down the plane
f=\(\mu\)*m2*g*cos30 -down the plane
resultant acc is up the plane and equal 0.2m/s/s, so
m2a =T-m2*g*sin30-\(\mu\)*m2*g*cos30 ...........................(2)

eqn 1 and 2 can be solved for 2 unknowns m2 and T

Answer 8

woah... now I am extremely confused.... what is the equation?

Answer 9

can u tell me the answer of this question so that i can see if what i did is correct.

Answer 10

I dont know what answer is.... i have an example one tho

Answer 11

Two Masses, a Pulley, and an Inclined Plane
Block 1, of mass m_1 = 0.700kg , is connected over an ideal (massless and frictionless) pulley to block 2, of mass m_2, as shown. For an angle of theta = 30.0^\circ and a coefficient of kinetic friction between block 2 and the plane of mu = 0.250, an acceleration of magnitude a = 0.350m/s^2 is observed for block 2.
Part A
Find the mass of block 2, m_2.
Express your answer numerically in kilograms.
Hint 1. Draw a free-body diagram
Which figure depicts the correct free-body diagrams for the blocks in this problem?
Figure a




Figure b




Figure c
The notation in the figures is as follows: T_vec is the tension in the string, n_vec is the normal force between block 2 and the surface of the incline, f_vec is the friction force, a_vec is the acceleration of the blocks, and g is the magnitude of the acceleration due to gravity.
ANSWER:
a
b
c
A
Hint 2. Apply Newton's 2nd law to block 2 in the direction parallel to the incline
Which of the following relations expresses Newton's 2nd law for block 2 in the direction parallel to the incline? (Assume the positive direction is going up the incline.)
Hint 1. Find the component of the weight of block 2 in the direction parallel to the incline
Which of the following expressions is the component of the weight of block 2 in the direction parallel to the incline? (Assume the positive direction is going up the incline.)
ANSWER:
-m_2 g \sin(30^\circ)
-m_2 g \cos(30^\circ)
m_2 g \sin(30^\circ)
m_2 g \cos(30^\circ)
A
Hint 2. Direction of the frictional force
Recall that the frictional force always acts in the direction opposite to the direction of motion.
ANSWER:
m_2 a=T-f-m_2 g\cos(30^\circ)
m_2 a=T-f-m_2 g\sin(30^\circ)
m_2 a=T+f-m_2 g
m_2 a=T-f+m_2 g\sin(30^\circ)
m_2 a=T-f-m_2 g
m_2 a=T-f+m_2 g\cos(30^\circ)
B
Hint 3. Find the tension
Find the magnitude T of the tension in the string.
Express your answer in newtons.
Hint 1. Apply Newton's 2nd law to block 1 in the vertical direction
Which of the following relations expresses Newton's 2nd law in the vertical direction for block 1? Take the positive direction to point downward.
ANSWER:
m_1 a =m_1 g+T
m_1 a =T-m_1 g
m_1 a =m_1 g-T
m_1 a =m_1 g-2T
m_1 a =m_1 g
m_1 a = -T
C
ANSWER:
T =
6.62
\rm N
Hint 4. Find an expression for the friction force
What is the magnitude f of the friction force acting on block 2?
ANSWER:
f=n/\mu
f=\mu/n
f=\mu n
\vec{f}=\mu\vec{n}
C
Hint 5. Find the normal force
By applying Newton's 2nd law to block 2 in the direction perpendicular to the incline determine the magnitude of the normal force n.
Express your answer in terms of m_2.
Hint 1. Find the component of the weight of block 2 in the direction perpendicular to the incline
Which of the following expressions is the component of the weight of block 2 in the direction perpendicular to the incline? (Assume the positive direction is going upward.)
ANSWER:
-m_2 g \sin(30^\circ)
-m_2 g \cos(30^\circ)
m_2 g \sin(30^\circ)
m_2 g \cos(30^\circ)
B
ANSWER:
n =
8.50
m_2
ANSWER:
m_2 =
0.897
\rm kg

Answer 12

did that help?

Answer 13

When i solve eqn 1 and 2 in my solution above for the ratio of m1 and m2, i get the same answer that u posted in your second reply.

And after plugging the values in that relation, i am getting m2~1kg

Answer 14

so 1 kg, is for which one? the first question i posted or the second? and what equation did u use to get that?

Answer 15

nvm, figured it out

Answer 16

Apply N's 2nd law to mass 1: it will reveal the tension in the string.

Then apply it to mass 2 knowing tension and acceleration. The only unknown will be mass m2.